Depression of a freezing point of the solutions depends on the number of particles of the solute in the solution.
1 mol of C6H12O6 after dissolving in water still be 1 mol, because C6H12O6 does no dissociate in water.
1 mol of C2H5OH after dissolving in water still be 1 mol, because C2H5OH does no dissociate in water.
1 mol of NaCl after dissolving in water gives 2 mol of particles (ions), because NaCl is a strong electrolyte(as salt) and completely dissociates in water.
NaCl ----->Na⁺ + Cl⁻
1 mol of CH3COOH after dissolving in water gives more than 1 mol but less than 2 moles, because CH3COOH is a weak electrolyte (weak acid) and dissociates only partially.
So, most particles of the solute is going to be in the solution of NaCl,
so<span> the lowest freezing point has the aqueous solution of NaCl.</span>
The net amount of energy produced can be obtained from a table of enthalpy change of formation, available online.
The enthalpy change of formation indicate how much energy the 1 mole of the product (H2O) has relative to the elemental reactants (H2 and O2). In other words, the "lost" energy equals the heat/energy released.
For water (H2O), this value is -285.8 if the final product is a liquid under standard conditions, and -241.82 if the product is in gas form which contains some energy that could be further released. This means that if the final product (H2O) is in liquid form, energy released is 285.8 kJ/mol.
Since water is in liquid form under standard conditions, the first value (285.8 kJ/mol) is generally appropriate.
Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, 
Explanation : Given,
Wavelength = 
conversion used : 
Formula used :

As, 
So, 
where,
= frequency
h = Planck's constant = 
= wavelength = 
c = speed of light = 
Now put all the given values in the above formula, we get:


Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, 
Answer:
[Na₂CO₃] = 0.094M
Explanation:
Based on the reaction:
HCO₃⁻(aq) + H₂O(l) ↔ CO₃²⁻(aq) + H₃O⁺(aq)
It is possible to find pH using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A⁻] / [HA]
Where [A⁻] is concentration of conjugate base, [CO₃²⁻] = [Na₂CO₃] and [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.
pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>
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Replacing these values:
10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]
<em> [Na₂CO₃] = 0.094M</em>
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